Problem: The equation of a circle $C$ is $x^2+y^2+18x-6y+54 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+18x) + (y^2-6y) = -54$ $(x^2+18x+81) + (y^2-6y+9) = -54 + 81 + 9$ $(x+9)^{2} + (y-3)^{2} = 36 = 6^2$ Thus, $(h, k) = (-9, 3)$ and $r = 6$.